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zepelin [54]
2 years ago
15

Mary determined that 5 pollen grains could fit along the diameter of the field of view of her microscope. If each pollen grain h

as a diameter of 0.3mm what is the diameter in millimeters of the field of view of Mary's microscope
1.5mm
15mm
4.7mm
5.3mm
Biology
1 answer:
KengaRu [80]2 years ago
7 0

Answer:

1.5mm

Explanation:

According to this question, Mary is trying to view 5 pollen grains under her microscope. She found out that the 5 pollen grains each with a diameter of 0.3mm could fit along the diameter of the field of view of her microscope.

This means that the diameter of the field of view of Mary's microscope can be calculated as follows:

Diameter of each pollen × no. of pollen that fit in

= 0.3mm × 5

= 1.5mm

Hence, the diameter of the field of view of Mary's microscope is 1.5mm.

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Answer:

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b. positive charge.

c. As atomic number is equal to no. of protons and no. of electron and atomic number 10 element is Neon.

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2a. 4 protons

b. 2 electrons

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3 years ago
Why are the shrubs responsible for<br> migrating animals?
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Explanation:

I found this information from https://jeb.biologists.org/content/222/Suppl_1/jeb191890

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3 years ago
(c) Based on the data in Table 1, describe why the dominant alleles for body color and wing shape are the alleles that produce a
Brrunno [24]

Answer:

Check the explanation

Explanation:

Consider the below table of F2 data of phenotypic flies: we can reconstruct the table as seen in the second attached image below:

From the above observations it is clear the Gray body long wings and Ebony body vestigial wings are the parental combinations and that the Gray body vestigial wings and Ebony body long wings are the recombinants.

Also the ratio indicates that Gray body is dominant over ebony body and that long wings is dominant over vestigial wings.

here the genes for the two characters have shown independent assortment which means that the genes are unlinked if located on the same chromosome or are located  on different chromosomes.

Now F1 hybrid= GgLl (G for Grey and L for Long)

Cross between F1 hybrid and true breeding Gray vestigial (GGll)

GgLl x GG ll

Gametes-----------> GL Gl gL gl Gl

        GL                          Gl                      gL                  gl

Gl    GGLl                      GGll                  GgLl               Ggll

   (Gray long)      (Gray vestigial)     (gray Long)     (Gray vestigial)

Therefore the probability of getting the flies with gray body vestigial wings= 2/4= 50%

b) The reason why the students obtained the above given F2 results involving a cross between true breeding Gray body and long wings with true breeding Ebony body and vestigial wings is that the genes for the two charcters asssort independently in F2 generation and that the genes are not linked as:

Parents------------------> GGLL x ggll

Gametes -----------------> GL gl

F1---------------------> GgLl (Gray long but in heterozous condition)

Now GgLl x GgLl

Gametes GL Gl gL gl   GL Gl gL gl

Here gametes assorted independently and hence in F2 generation we got the above results (U can show the results in the form of punnett square.

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